Answer:
literally same im failing almost every class and i just know my dad hates me because of it
Explanation:
<h2>a)
The rate at which
is formed is 0.066 M/s</h2><h2>b)
The rate at which molecular oxygen
is reacting is 0.033 M/s</h2>
Explanation:
Rate law says that rate of a reaction is directly proportional to the concentration of the reactants each raised to a stoichiometric coefficient determined experimentally called as order.

The rate in terms of reactants is given as negative as the concentration of reactants is decreasing with time whereas the rate in terms of products is given as positive as the concentration of products is increasing with time.
Rate in terms of disappearance of
=
= 0.066 M/s
Rate in terms of disappearance of
= ![-\frac{1d[O_2]}{dt}](https://tex.z-dn.net/?f=-%5Cfrac%7B1d%5BO_2%5D%7D%7Bdt%7D)
Rate in terms of appearance of
= ![\frac{1d[NO_2]}{2dt}](https://tex.z-dn.net/?f=%5Cfrac%7B1d%5BNO_2%5D%7D%7B2dt%7D)
1. The rate of formation of 
![-\frac{d[NO_2]}{2dt}=\frac{1d[NO]}{2dt}](https://tex.z-dn.net/?f=-%5Cfrac%7Bd%5BNO_2%5D%7D%7B2dt%7D%3D%5Cfrac%7B1d%5BNO%5D%7D%7B2dt%7D)
![\frac{1d[NO_2]}{dt}=\frac{2}{2}\times 0.066M/s=0.066M/s](https://tex.z-dn.net/?f=%5Cfrac%7B1d%5BNO_2%5D%7D%7Bdt%7D%3D%5Cfrac%7B2%7D%7B2%7D%5Ctimes%200.066M%2Fs%3D0.066M%2Fs)
2. The rate of disappearance of 
![-\frac{1d[O_2]}{dt}=\frac{d[NO]}{2dt}](https://tex.z-dn.net/?f=-%5Cfrac%7B1d%5BO_2%5D%7D%7Bdt%7D%3D%5Cfrac%7Bd%5BNO%5D%7D%7B2dt%7D)
![-\frac{1d[O_2]}{dt}=\frac{1}{2}\times 0.066M/s=0.033M/s](https://tex.z-dn.net/?f=-%5Cfrac%7B1d%5BO_2%5D%7D%7Bdt%7D%3D%5Cfrac%7B1%7D%7B2%7D%5Ctimes%200.066M%2Fs%3D0.033M%2Fs)
Learn more about rate law
brainly.com/question/13019661
https://brainly.in/question/1297322
Answer:
The concentration of I at equilibrium = 3.3166×10⁻² M
Explanation:
For the equilibrium reaction,
I₂ (g) ⇄ 2I (g)
The expression for Kc for the reaction is:
![K_c=\frac {\left[I_{Equilibrium} \right]^2}{\left[I_2_{Equilibrium} \right]}](https://tex.z-dn.net/?f=K_c%3D%5Cfrac%20%7B%5Cleft%5BI_%7BEquilibrium%7D%20%5Cright%5D%5E2%7D%7B%5Cleft%5BI_2_%7BEquilibrium%7D%20%5Cright%5D%7D)
Given:
= 0.10 M
Kc = 0.011
Applying in the above formula to find the equilibrium concentration of I as:
![0.011=\frac {\left[I_{Equilibrium} \right]^2}{0.10}](https://tex.z-dn.net/?f=0.011%3D%5Cfrac%20%7B%5Cleft%5BI_%7BEquilibrium%7D%20%5Cright%5D%5E2%7D%7B0.10%7D)
So,
![\left[I_{Equilibrium} \right]^2=0.011\times 0.10](https://tex.z-dn.net/?f=%5Cleft%5BI_%7BEquilibrium%7D%20%5Cright%5D%5E2%3D0.011%5Ctimes%200.10)
![\left[I_{Equilibrium} \right]^2=0.0011](https://tex.z-dn.net/?f=%5Cleft%5BI_%7BEquilibrium%7D%20%5Cright%5D%5E2%3D0.0011)
![\left[I_{Equilibrium} \right]=3.3166\times 10^{-2}\ M](https://tex.z-dn.net/?f=%5Cleft%5BI_%7BEquilibrium%7D%20%5Cright%5D%3D3.3166%5Ctimes%2010%5E%7B-2%7D%5C%20M)
<u>Thus, The concentration of I at equilibrium = 3.3166×10⁻² M</u>
Answer: The derivative of a constant term is always 0. So the acceleration of the body would be zero. Hence, the acceleration of a body moving with uniform velocity will always be zero.
Hope this helps :) :)
Actually, that does not happen until the protostar becomes a star when nuclear ignition starts and is maintained. It takes awhile for new star to go through its T-Tauri stage and settle down on the main sequence.
<span>A STAR does not reach hydrostatic equilibrium until it on the main sequence. Otherwise, it would remain a brown dwarf with not enough mass to to maintain nuclear fusion for more than 3,000 to 10,00 years. </span>